Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{x + 9}{-7x - 42} \times \dfrac{3x - 3}{x^2 + 8x - 9} $
Answer: First factor the quadratic. $z = \dfrac{x + 9}{-7x - 42} \times \dfrac{3x - 3}{(x + 9)(x - 1)} $ Then factor out any other terms. $z = \dfrac{x + 9}{-7(x + 6)} \times \dfrac{3(x - 1)}{(x + 9)(x - 1)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (x + 9) \times 3(x - 1) } { -7(x + 6) \times (x + 9)(x - 1) } $ $z = \dfrac{ 3(x + 9)(x - 1)}{ -7(x + 6)(x + 9)(x - 1)} $ Notice that $(x - 1)$ and $(x + 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ 3\cancel{(x + 9)}(x - 1)}{ -7(x + 6)\cancel{(x + 9)}(x - 1)} $ We are dividing by $x + 9$ , so $x + 9 \neq 0$ Therefore, $x \neq -9$ $z = \dfrac{ 3\cancel{(x + 9)}\cancel{(x - 1)}}{ -7(x + 6)\cancel{(x + 9)}\cancel{(x - 1)}} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $z = \dfrac{3}{-7(x + 6)} $ $z = \dfrac{-3}{7(x + 6)} ; \space x \neq -9 ; \space x \neq 1 $